Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

It a hard problem, but I know how to solve it (I just learned binomial formula)

Ok the first step is to find combinations 5C3= 5!/(5-3)!*3!=5*4/2=10
Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8)
Third is to find the probability of all unwanted events, 1/2^2 (1/4)

now you have to multiply all these numbers

10/8*4

the answer is 10/32 or 5/16

P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)

Hi you answer would have been correct if the problem would have been "Three Heads" but I think you overlooked Consecutive heads...so the asnwer is 3/16.

Saurabh Malpani

Sergey_is_cool wrote:

It a hard problem, but I know how to solve it (I just learned binomial formula)

Ok the first step is to find combinations 5C3= 5!/(5-3)!*3!=5*4/2=10 Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8) Third is to find the probability of all unwanted events, 1/2^2 (1/4)

now you have to multiply all these numbers

10/8*4

the answer is 10/32 or 5/16

P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)

Hi you answer would have been correct if the problem would have been "Three Heads" but I think you overlooked Consecutive heads...so the asnwer is 3/16.

Saurabh Malpani

Sergey_is_cool wrote:

It a hard problem, but I know how to solve it (I just learned binomial formula)

Ok the first step is to find combinations 5C3= 5!/(5-3)!*3!=5*4/2=10 Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8) Third is to find the probability of all unwanted events, 1/2^2 (1/4)

now you have to multiply all these numbers

10/8*4

the answer is 10/32 or 5/16

P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)

The question stem asks not just about consecutive heads, but ''at least three'' consecutive heads: this means, of course, the sum of the probobility of 3 consecutive heads, the probability of 4 consecutive heads, and the probability of 5 consecutive heads. So--Himalayan-- you have these possibilies: HHHTT, THHHT, TTHHH, HHHHT, THHHH, and HHHHH. This ends up resulting in a probability of 6/32 = 3/16.